cp's OEIS Frontend

Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001

Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_8. Example: a(1)=3 because in the cycle ABCDEFGH we have three walks of length 3 between A and B: ABAB, ABCB and AHAB. - Emeric Deutsch, Apr 01 2004

Smallest number containing in its binary representation two equal non-overlapping subwords of length n: A097295(a(n))=n and A097295(m)Reinhard Zumkeller, Aug 04 2004

a(n)^2 + (A006516(n))^2 = a(2n). E.g., a(3) = 36, A006516(3) = 28, a(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006

Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either x equals y or x does not equal y. - Ross La Haye, Jan 02 2008

Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A). This is just a simpler statement of my previous comment for this sequence. - Ross La Haye, Jan 10 2008

For n>0: A000120(a(n))=2, A023414(a(n))=2*(n-1), A087117(a(n))=n-1. - Reinhard Zumkeller, Jun 23 2009

a(n+1) written in base 2: 11, 1010, 100100, 10001000, 1000010000, ..., i.e., number 1, n times 0, number 1, n times 0 (A163449(n)). - Jaroslav Krizek, Jul 27 2009

a(n) for n >= 1 is a bisection of A001445(n+1). - Jaroslav Krizek, Aug 14 2009

Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times sum_{k=0..n} binomial(n,k)*k^q, then A007582(x)= sum_{k=0..x-1} T(x,k)*2^k. - John M. Campbell, Nov 16 2011

a(n) gives the number of pairs (r, s) such that 0 <= r <= s <= (2^n)-1 that satisfy AND(r, s, XOR(r, s)) = 0. - Ramasamy Chandramouli, Aug 30 2012

a(n) = A000217(2^n) = 2^(2n-1) + 2^(n-1) is the nearest triangular number above 2^(2n-1); cf. A006516, A233327. - Antti Karttunen, Feb 26 2014

Consider the quantum spin-1/2 chain with even number of sites L (physics, condensed matter theory). The spectrum of the Hamiltonian can be classified according to symmetries. If the only symmetry of the spin Hamiltonian is Parity, i.e., reflection with respect to the middle of the chain (see e.g. the transverse-field Ising model with open boundary conditions), then the dimension of the p=+1 parity sector is given by a(n) with n=L/2. - Marin Bukov, Mar 11 2016

a(n) is also the total number of words of length n, over an alphabet of four letters, of which one of them appears an even number of times. See the Lekraj Beedassy, Jul 22 2003, comment on A006516 (4-letter odd case), and the Balakrishnan reference there. For the 1- to 11-letter cases, see the crossrefs. - Wolfdieter Lang, Jul 17 2017

a(n) is the number of nonisomorphic spanning trees of the cyclic snake formed with n+1 copies of the cycle on 4 vertices. A cyclic snake is a connected graph whose block-cutpoint is a path and all its n blocks are isomorphic to the cycle C_m. - Christian Barrientos, Sep 05 2024

Also, with offset 1, the cogrowth sequence of the dihedral group with 16 elements, D8 = . - Sean A. Irvine, Nov 06 2024

The sequence divides naturally into blocks of length 2^k, k = 0, 1, 2, ... On block k, let n go from 0 to 2^k-1, write n in binary using k bits and reverse the bits. - N. J. A. Sloane, Jun 11 2002

For example: the 3-bit strings are 000, 001, 010, 011, 100, 101, 110 and 111. When they are bit-reversed, we get 000, 100, 010, 110, 001, 101, 011, 111. Or, in decimal representation 0,4,2,6,1,5,3,7.

In other words: Given any n>1, the set of numbers A030109[i] for indexes i ranging from 2^n to 2^(n+1)-1 is a permutation of the set of consecutive integers {0,1,2,...,2^n-1}. Example: for n=2, we have the permutation of {0,2,1,3} of {0,1,2,3} This is important in the standard FFT algorithms requiring a starting (or ending) bit-reversal permutation of indices. - Stanislav Sykora, Mar 15 2012

Name when submitted: Sum of even-indexed terms of n-th row of array T given by A049773 (from Clark Kimberling).

Also sum of integers of which the binary order [A029837] is n: a(n) = Sum_[x | ceiling(log_2(x)) = n ]. E.g., a(7) = 6176 = Apply[Plus, Table[w,{w,65,128}]].

This sequence may be obtained by filling a complete binary tree left-to-right, row by row with the integers onwards from 2 and then collecting the sums of the rows; e.g., 2, 3+4, 5+6+7+8, 9+10+11+12+13+14+15+16, etc. a(n) is then equal to the sum of row n-1. - Carl R. White, Aug 19 2003

If the offset is set to zero, the inverse binomial transform gives A007051 without its leading 1. - R. J. Mathar, Mar 26 2009

Schroeder, p. 281 states "The ordering with which the iterates x_n fall into the 2^m different chaos bands [order as to magnitude] is also the same as the ordering of the iterates in a stable orbit of period length P = 2^m. For example, for both the period-4 orbit and the four chaos bands, the iterates, starting with the largest iterate x_1, are ordered as follows: x_1 > x_3 > x_4 > x_2."

From Andrey Zabolotskiy, Dec 06 2024: (Start)

For n>0, row n-1 is the permutation relating row n of the left half of Stern-Brocot tree with row n of Kepler's tree of fractions. Specifically, if K_n(k) [resp. SB_n(k)] is the k-th fraction in the n-th row of A294442 [resp. A057432], where 1/2 is in row 1 and k=1..2^(n-1), then SB_n(k) = K_n(T(n-1, k)).

The inverse permutation is row n of A131271.

Equals A362160+1. (End)

The n-th row differs from the prior row only by the presence of n. See A088371 for the positions in the n-th row that n is inserted.

From Clark Kimberling, Aug 02 2007: (Start)

At A131966, this sequence is cited as the fractal sequence of the Cantor set C.

Recall that C is the set of fractions in [0,1] whose base 3 representation consists solely of 0's and 2's.

Arrange these fractions as follows:

0

0, .2

0, .02, .2

0, .02, .2, .22

0, .002, .02, .2, .22, etc.

Replace each number x by its order of appearance, counting each distinct predecessor of x only once, getting

1;

1, 2;

1, 3, 2;

1, 3, 2, 4;

1, 5, 3, 2, 4;

Concatenate these to get the current sequence, which is a fractal sequence as defined in "Fractal sequences and interspersions".

One property of such a sequence is that it properly contains itself as a subsequence (infinitely many times). (End)

Row n contains one of A003407(n) non-averaging permutations of [n], i.e., a permutation of [n] without 3-term arithmetic progressions. - Alois P. Heinz, Dec 05 2017

In a knockout competition with 2^n players, arranging the competition brackets (see Wikipedia) in T(n,k) order, where T(n,k) is the rank of the k-th player, ensures that highest ranked players cannot meet until the later stages of the competition. None of the top 2^p ranked players can meet earlier than the p-th from last round of the competition. At the same time the top ranked players in each match meet the lowest ranked player possible consistent with this rule. The sequence for the top ranked players meeting the highest ranked player possible is A049773. - Colin Hall, Feb 28 2012

Ranks in natural order of 2^n increasing real numbers appearing in limit cycles of interval iterations, or Median Spiral Order. [See the works by Demongeot & Waku]

From Andrey Zabolotskiy, Dec 06 2024 (Start):

For n>0, row n-1 is the permutation relating row n of Kepler's tree of fractions with row n of the left half of Stern-Brocot tree. Specifically, if K_n(k) [resp. SB_n(k)] is the k-th fraction in the n-th row of A294442 [resp. A057432], where 1/2 is in row 1 and k=1..2^(n-1), then K_n(k) = SB_n(T(n-1, k)).

The inverse permutation is row n of A088208.

When 1 is subtracted from each term, rows 3-5 become A240908, A240909, A240910. (End)

Define binary weight wt(n) as A000120(n), the number of 1s in the binary expansion of n. Let w = A000120(a(n-1)) the binary weight of the previous term. In other words, a(n) is the least m not already in the sequence such that m mod 2^k = w, where k = floor(1 + log_2 w).

Likely a permutation of the natural numbers.

The numbers m = 2^k with 0 <= k <= 3 appear at indices {1, 3, 11, 222}. The term 16 has not appeared for n <= 2^14 and may not until n approaches 2^16.

The numbers m = (2^k + 1) appear at indices {2, 4, 12, 223, ...}. The numbers m = 2^k or (2^k + 1) require n approximately equal to 2^m in order to appear in the sequence.

The numbers m = (2^k - 1) with 1 <= k <= 14 appear at indices {1, 2, 8, 10, 23, 43, 130, 278, 447, 758, 1390, 2525, 4719, 9333}, respectively.

The plot exhibits dendritic streams of residues r (mod 2^k). We can identify coordinates (x, y) = (n, a(n)) on the plot where the streams branch.

The branches of the tree in the plot contain m congruent to r (mod 2^k), where r is a term (except the last term) in row (k-1) of A049773.

Given 2^14 terms of this sequence, we see 2 or 3 successive invocations of w, otherwise, w appears just once before a different value succeeds it in the next term.

2^4 appears at index 47201. - Michael S. Branicky, Dec 16 2020

A permutation of the integers since n appears at or before index 2^n - 1, the first number with binary weight n. - Michael S. Branicky, Dec 16 2020

1

1..............................................2

1......................3.......................2.......................4

1...........5..........3...........7...........2...........6...........4...........8

1.....9.....5....13....3....11.....7....15.....2....10.....6....14.....4....12.....8....16

1.17..9.25..5.21.13.29.3.19.11.27..7.23.15.31..2.18.10.26..6.22.14.30..4.20.12.28..8.24.16.32

...

To obtain row n (with n > 0):

- take a strip of paper of dimensions 2^(n-1) X 1

- number the square cells from left to right from 1 to 2^(n-1)

- fold this strip of paper n-1 times, in the middle, covering the left part with the right part; at the end all the cells are stacked on the cell with the number 1

- read the numbers written on square cells from bottom to top.

For n > 0:

- T(n,1) = 1 (the first cell always stays at the bottom)

- T(n+1,2) = 2^n (the last cell covers the first cell after the first folding)

- T(n+1,2^n) = 2 (the second cell comes on top after the last folding).

For n > 0 and k=1..2^(n-2):

- T(n+1,2*k-1) + T(n+1,2*k) = 2^n+1 (opposite cells (summing to 2^n+1) are paired after the first folding).

This sequence has similarities with A049773: here we fold in the middle; there we cut in the middle, covering the left part with the right part.